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Let $\omega$ denote the set of natural numbers, let $\text{Sym}(\omega)$ be the collection of bijections $\psi:\omega\to\omega$, and let $\text{(fin)}$ be the set of members of $\text{Sym}(\omega)$ having finite support. Formally, $$\text{(fin)} = \{\psi\in\text{Sym}(\omega): (\exists k\in\omega)(\forall i\in\omega\setminus k)\psi(i) = i\}.$$

Is $\text{Sym}(\omega)/\text{(fin)}$ isomorphic to a subgroup of $\text{Sym}(\omega)$?

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    $\begingroup$ Perhaps we can somehow use the fact that every collection of disjoint sets in $\omega$ is countable, but there are uncountable collections of sets that are disjoint mod finite. $\endgroup$
    – Joel David Hamkins
    Nov 26 at 16:25
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    $\begingroup$ Exact duplicate of your own question! $\endgroup$
    – YCor
    Nov 26 at 18:11
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    $\begingroup$ @JoelDavidHamkins yes that's exactly what's used in the original answer to the question, implying that there's a subset $X$ in this group whose centralizer is not equal to the centralizer of any countable subset of $X$. This implies that this group doesn't embed into any group with a metrizable separable topology. $\endgroup$
    – YCor
    Nov 26 at 19:53
  • $\begingroup$ Right -- apologies for the duplicate $\endgroup$
    – Dominic van der Zypen
    Nov 27 at 7:19

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$Sym(\omega)$ has what is called the "automatic continuity property", whence any morphism from $Sym(\omega)$ to a Polish group $G$ which is trivial on $(fin)$ is the identity (by density of $(fin)$ and continuity). Hence the only map from $Sym(\omega)/(fin)$ to $Sym(\omega)$ is the trivial map.

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  • $\begingroup$ Surely you mean "is trivial" rather than "is the identity"? $\endgroup$
    – Yemon Choi
    Nov 26 at 19:02
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    $\begingroup$ @YemonChoi There is a natural topology whose basic open sets are the permutations extending some finite partial map. This is an analogue of the product topology, and the finitely supported permutations are dense, since any finite injective map can be extended to a finitely supported permutation. $\endgroup$
    – Joel David Hamkins
    Nov 26 at 19:39
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    $\begingroup$ Thanks @JoelDavidHamkins $\endgroup$
    – Yemon Choi
    Nov 26 at 20:13
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    $\begingroup$ @YemonChoi it's just the pointwise convergence topology (i.e., it is equal, and not just an analogue, to the topology induced by the product topology of the countable product $\omega^\omega$). $\endgroup$
    – YCor
    Nov 27 at 6:33
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    $\begingroup$ Maybe this answer should be posted on the original question too (maybe with an extra word or reference about why $\mathrm{Sym}(\omega)$ has the automatic continuity property?), since it is substantially different from the answer given there. $\endgroup$
    – Sean Eberhard
    Nov 27 at 10:08

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